Answer

Verified

448.5k+ views

Hint: First of all assume the time taken by a larger pipe to fill the pool as ‘T’ hours. From this, we will get the time for a smaller pipe as well. Then find the capacity of the pool filled by each of the pipes in one hour. Then use,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{1}{2}\] where ‘L’ is the total capacity of the pool.

Complete step-by-step answer:

Here we are given two pipes, one of larger diameter and other of smaller diameter which takes 4 hours and 9 hours respectively together to fill half of the pool. Also, we are given that pipe of the smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. We have to find the time taken by each pipe to separately fill the pool.

Let us consider the total capacity of the pool to be ‘L’ liters.

Also, let us assume that a pipe of larger diameter takes ‘T’ hours to fill this pool.

Since this pipe takes ‘T’ hours to fill ‘L’ liters. Therefore, in 1 hour, it would fill \[=\dfrac{L}{T}\text{liters}\].

So, in 4 hours this pipe of larger diameter would fill \[=4\times \dfrac{L}{T}\text{liters}.....\left( i \right)\]

Now, we are also given that the pipe of the smaller diameter takes 10 more hours to fill the pool as compared to the pipe of the larger diameter. So, we get,

The time taken by a smaller pipe to fill the pool = (T + 10) hours.

Since this pipe takes ‘(T + 10)’ hours to fill ‘L’ liters, therefore in 1 hour, it would fill \[=\dfrac{L}{\left( T+10 \right)}\text{liters}\].

So, in 9 hours this pipe of smaller diameter would fill \[=9\times \dfrac{L}{\left( T+10 \right)}\text{liters}....\left( ii \right)\]

Now, we are given that pipe of larger diameter in 4 hours and pipe of smaller diameter in 9 hours together fills half of the pool, that is, \[\dfrac{L}{2}\text{liters}\]. So, we get,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{L}{2}\]

By substituting the values from equation (i) and (ii), we get,

\[\dfrac{4L}{T}+\dfrac{9L}{T+10}=\dfrac{L}{2}\]

By taking out ‘L’ common from both the sides and canceling it, we get,

\[\dfrac{4}{T}+\dfrac{9}{T+10}=\dfrac{1}{2}\]

By simplifying the above equation, we get,

\[\dfrac{4\left( T+10 \right)+9T}{T\left( T+10 \right)}=\dfrac{1}{2}\]

By cross multiplying the above equation, we get,

\[2\left( 4T+40+9T \right)={{T}^{2}}+10T\]

\[\Rightarrow 2\left( 13T+40 \right)={{T}^{2}}+10T\]

\[\Rightarrow 26T+80={{T}^{2}}+10T\]

By taking all the terms to one side, we get,

\[\Rightarrow {{T}^{2}}+10T-26T-80=0\]

Or, \[{{T}^{2}}-16T-80=0\]

Here, we can write \[16T=20T-4T\], so we get,

\[\Rightarrow {{T}^{2}}-20T+4T-80=0\]

We can also write the above equation as,

\[T\left( T-20 \right)+4\left( T-20 \right)=0\]

By taking out (T – 20) common, we get,

\[\left( T-20 \right)\left( T+4 \right)=0\]

We get, T = 20 and T = – 4

Since we know that the value of T can’t be negative, hence we get T = 20 hours.

Therefore, we get the total time taken by the pipe of larger diameter to fill the pool is equal to 20 hours.

Also, we get the total time taken by the pipe of smaller diameter to fill the pool is equal to (T + 10) = 30 hours.

Note: In these types of questions, students should always use the approach of the unitary method to solve the questions, that is, always calculate for 1 unit and they multiply it with the number of units you are asked for. Also, students can cross-check their answer by substituting the value of T = 20 hours back in the equation and checking if LHS = RHS or not.

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{1}{2}\] where ‘L’ is the total capacity of the pool.

Complete step-by-step answer:

Here we are given two pipes, one of larger diameter and other of smaller diameter which takes 4 hours and 9 hours respectively together to fill half of the pool. Also, we are given that pipe of the smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool. We have to find the time taken by each pipe to separately fill the pool.

Let us consider the total capacity of the pool to be ‘L’ liters.

Also, let us assume that a pipe of larger diameter takes ‘T’ hours to fill this pool.

Since this pipe takes ‘T’ hours to fill ‘L’ liters. Therefore, in 1 hour, it would fill \[=\dfrac{L}{T}\text{liters}\].

So, in 4 hours this pipe of larger diameter would fill \[=4\times \dfrac{L}{T}\text{liters}.....\left( i \right)\]

Now, we are also given that the pipe of the smaller diameter takes 10 more hours to fill the pool as compared to the pipe of the larger diameter. So, we get,

The time taken by a smaller pipe to fill the pool = (T + 10) hours.

Since this pipe takes ‘(T + 10)’ hours to fill ‘L’ liters, therefore in 1 hour, it would fill \[=\dfrac{L}{\left( T+10 \right)}\text{liters}\].

So, in 9 hours this pipe of smaller diameter would fill \[=9\times \dfrac{L}{\left( T+10 \right)}\text{liters}....\left( ii \right)\]

Now, we are given that pipe of larger diameter in 4 hours and pipe of smaller diameter in 9 hours together fills half of the pool, that is, \[\dfrac{L}{2}\text{liters}\]. So, we get,

(Pool filled by the pipe of larger diameter in 4 hours) + (Pool filled by the pipe of smaller diameter in 9 hours) = \[\dfrac{L}{2}\]

By substituting the values from equation (i) and (ii), we get,

\[\dfrac{4L}{T}+\dfrac{9L}{T+10}=\dfrac{L}{2}\]

By taking out ‘L’ common from both the sides and canceling it, we get,

\[\dfrac{4}{T}+\dfrac{9}{T+10}=\dfrac{1}{2}\]

By simplifying the above equation, we get,

\[\dfrac{4\left( T+10 \right)+9T}{T\left( T+10 \right)}=\dfrac{1}{2}\]

By cross multiplying the above equation, we get,

\[2\left( 4T+40+9T \right)={{T}^{2}}+10T\]

\[\Rightarrow 2\left( 13T+40 \right)={{T}^{2}}+10T\]

\[\Rightarrow 26T+80={{T}^{2}}+10T\]

By taking all the terms to one side, we get,

\[\Rightarrow {{T}^{2}}+10T-26T-80=0\]

Or, \[{{T}^{2}}-16T-80=0\]

Here, we can write \[16T=20T-4T\], so we get,

\[\Rightarrow {{T}^{2}}-20T+4T-80=0\]

We can also write the above equation as,

\[T\left( T-20 \right)+4\left( T-20 \right)=0\]

By taking out (T – 20) common, we get,

\[\left( T-20 \right)\left( T+4 \right)=0\]

We get, T = 20 and T = – 4

Since we know that the value of T can’t be negative, hence we get T = 20 hours.

Therefore, we get the total time taken by the pipe of larger diameter to fill the pool is equal to 20 hours.

Also, we get the total time taken by the pipe of smaller diameter to fill the pool is equal to (T + 10) = 30 hours.

Note: In these types of questions, students should always use the approach of the unitary method to solve the questions, that is, always calculate for 1 unit and they multiply it with the number of units you are asked for. Also, students can cross-check their answer by substituting the value of T = 20 hours back in the equation and checking if LHS = RHS or not.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE